A calorimeter of water equivalent 20g contains 180g of water

Question:

A calorimeter of water equivalent $20 \mathrm{~g}$ contains $180 \mathrm{~g}$ of water at $25^{\circ} \mathrm{C}$. 'm' grams of steam at $100^{\circ} \mathrm{C}$ is mixed in it till the temperature of the mixure is $31^{\circ} \mathrm{C}$. The value of ' $\mathrm{m}$ ' is close to (Latent heat of water $=540 \mathrm{cal} \mathrm{g}^{-1}$, specific heat of water $=1 \mathrm{cal} \mathrm{g}^{-1}{ }^{\circ} \mathrm{C}^{-1}$ )

  1. $2.6$

  2.  2

  3. 4

  4. $3.2$


Correct Option: , 2

Solution:

$\frac{\mathrm{Cal}}{20 \mathrm{gm}} \frac{\mathrm{H}_{2} \mathrm{O}}{180 \mathrm{gm}} \frac{\text { Sterm }}{\mathrm{m}}$

$25^{\circ} \mathrm{C} \quad 25^{\circ} \mathrm{C} \quad 100^{\circ} \mathrm{C}$

$200 \times 1 \times(31-25)$

$=m \times 540+m \times 1 \times(100-31)$

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